in Linear Algebra recategorized by
5,798 views
31 votes

The eigen vector $(s)$ of the matrix $$\begin{bmatrix} 0 &0 &\alpha\\ 0 &0 &0\\ 0 &0 &0 \end{bmatrix},\alpha \neq 0$$ is (are)

  1. $(0,0,\alpha)$
  2. $(\alpha,0,0)$
  3. $(0,0,1)$
  4. $(0,\alpha,0)$
in Linear Algebra recategorized by
5.8k views

7 Comments

I believe the answer should be only B. It is because if we look at the matrix it shows the transformed 3D space into a line where the z^ vector is transformed into a line replicating x axis. In that scenario the only vector that remains unchanged is (alpha, 0, 0) i.e option B.
0
could anyone give the complete procedure ?
0
Answer is b only why obtion d is consider here ?
0
Why d is included in answer?

Here we got eign values as 0,0,0.

Now to get eign vector take  λ=0 in

( A-λI) X =0 or AX=0 -----(i)

here X is eign vector in form of column matrix [X1,X2,X3].

If we multiply as per (i) we get equation as

X1*0+X2*0+ aX3=0

Now for LHS to be 0 , X3 must be 0 . Here X1 and X2 can be assumed any value . As per the given option a and c are wrong.hence b and d satisfy the case
32
what about (alpha,alpha,0) vectors? will they be eigen vectors too.I think they are.
0
Bind blowing explanation satyajeet sir
0
Yes they will be eigen [email protected] gow
0

Subscribe to GO Classes for GATE CSE 2022

6 Answers

32 votes
 
Best answer
Since, the given matrix is an upper triangular one, all eigenvalues are 0. And hence $A-\lambda I=A$

So the question asks

$\begin{bmatrix} 0 &0 &\alpha \\ 0&0 &0 \\ 0& 0& 0 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

what $x_1,x_2,x_3$ are suitable?

which means $x_1$times column1+$x_2$times column2+$x_3$ times column3=Zero Vector

Look carefully,if you take any linear combination of columns 1 and 2 and don't take column 3(means $x_3=0$), the result will be a zero vector.

So, if we have our X=$\begin{bmatrix} \beta\\ \gamma\\ 0 \end{bmatrix}$ where $\beta, \gamma$ are any scalar quantity,

but $x_3$ must be necessarily zero to get zero vector.

Hence, only option (B) and (D) satisfy.
selected by

7 Comments

The second part of the answer is difficult for me to grasp.
1
Which one?
0
From the statement "look carefully"
1
ohh that one. See from this example.

Suppose you have a system like

$\begin{bmatrix} 1 &3 &6 \\ 7 & 5& 9\\ 3 &8 & 2 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$

this is equivalent to

$x_1\begin{bmatrix} 1\\ 7\\ 3 \end{bmatrix}+x_2\begin{bmatrix} 3\\ 5\\ 8 \end{bmatrix}+x_3\begin{bmatrix} 6\\ 9\\ 2 \end{bmatrix}$

In the question given, you have to produce a zero vector as result. What suitable combinations of $x_1,x_2,x_3$ can you take?
13
Good Explanation Sir ;-)
1
@Ayush sir, it will be 0,0,0 right?
0
0
25 votes
Given matrix is an upper triangular matrix. Therefore diagonal elements of A are eigen values of A, i.e. λ=0,0,0.

Consider the characteristic equation $(A-λI)x=0$

$\Rightarrow$ $\begin{bmatrix} 0-\lambda & 0 & \alpha \\ 0 & 0-\lambda & 0\\ 0 & 0 & 0-\lambda \end{bmatrix}$ $\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}$ = $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

To find the eigen vector of A corresponding to $\lambda$
Put the eigen value of A in the characteristic equation

i.e. $\begin{bmatrix} 0 & 0 & \alpha \\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$ $\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}$ = $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

Now we have

$\alpha$$x_3$=0

$\Rightarrow$$x_3$ = 0

Therefore, any non zero vector with z-component as zero is an eigen vector.
Hence in the given options $B$ and $D$ are correct.
edited by

1 comment

Thank you for explaining :)
1
16 votes
Answer: $B, D$

Eigen values are: $0,0,0$

The eigen vector should satisfy the equation: $\alpha$z $=$ $0$
edited by

10 Comments

yes, I also think ans should be option B only.
0
how b and d?
0
here by normal approach we get in eigen vector x3=0

So, answer will be b) and d)
0
^ Will $-\lambda^3 = 0$ not be the characterastic equations for this matrix?
So how third value came $\alpha$?
0
yes (B) and (D) should be correct option!
0

@Manu Thakur

here we have the Basic variable as "z" and free variable as "x" and "y".

So, keeping x = t and y = s, and z = $\alpha$

Then vector corresponding to x is $[1, 0, 0]^T$

And vector corresponding to y is $[0, 1, 0]^T$

Then why it is not $[1, 0, 0]^T$ And $[0, 1, 0]^T$ instead of $[\alpha, 0 , 0]^T$ And $[0, \alpha, 0]^T$?

0
$\alpha$ or 1 doesnot matter

atleast u have to give answer, from options given
0
But it should be $[1,0,0]^T$ And $[0,1,0]^T$ even no option matches. ryt??
0
if B & D are true then A should also be true please check once?
0
Ok
0
1 vote

Guys this is an easy question, so let us not make it complex.

As most of the people have already explained that $|A-\lambda I|*eigenvector =\begin{bmatrix} 0 &0 &\alpha \\ 0 & 0 &0 \\ 0& 0 & 0 \end{bmatrix}$$\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}$

Now see that we are getting $0*x_{1} + 0*x_{2} + \alpha *x_{3}=0$ as the only useful equation.

Now put whatever the value of $x_{1}$ ,$x_{2}$ and $x_{3}$, we will get our eigenvectors, EXCEPT the cases where $\alpha$ has to be zero because in question it clearly mentioned that $\alpha \neq 0$. This condition will only fail when $x_{3}\neq 0.$ And therefore now you can match the options very easily.

0 votes

clearly matrix is upper triangular matrix and for upper triangular ,lower triangular matrix ,diagonal matrix and scaler matrix ,its diagonal element is eigen values.so here eigen value=0,0,0.

now we have to find eigen vector of this matrix (A-λ)X=0  where λ=0 so we get A matrix  [ 0 0 α] has rank=1  means we get two free variable (n-r) 

x1= β x2=γ x3=0 then eigen vector  (x1,x2,x3)=(β,γ,0)= β(1,0,0)+γ(0,1,0) so  B and  D  are  right answer .

 

0 votes

 

https://youtu.be/VqP2tREMvt0

Check @32:57

After solving the characteristic equation $\det(A – \lambda I_n)=0$ we get $\lambda=0,0,0$. Using this $\lambda=0$ in $A-\lambda I_n$ we have:

 

Here $C_1$ and $C_2$ are any real. We may even choose them as $C_1=C_2=\alpha$ to get options $B$ and $D$.

 

edited by
Answer:

Related questions

Ask
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true